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Exodus Advanced Communications

Transmission Line Equations

Transmission lines take on many forms in order to accommodate particular applications. All rely on the same basic components - two or more conductors separated by a dielectric (insulator). The physical configuration and properties of all the components determines the characteristic impedance, distortion, transmission speed, and loss.

See a discussion on transmission lines and coaxial connectors.

The following formulas are presented in a compact text format that can be copied and pasted into a spreadsheet or other application.

For the following equations, ε is the dielectric constant (ε = 1 for air)

Two Conductors in Parallel (Unbalanced)
Above Ground Plane

For D << d, h

Z0= (69/ε½) log10{(4h/d)[1+(2h/D)2]-½}

Parallel conductors above a ground plane - RF Cafe
 Single Conductor Above Ground Plane
 
For d << h

Z0= (138/ε½) log10(4h/d)

Single conductor above a ground plane - RF Cafe
Two Conductors in Parallel (Balanced)
Above Ground Plane

For D << d, h1, h2

Z0= (276/ε½) log10{(2D/d)[1+(D/2h)2]-½}

Parallel conductors above a ground plane - RF Cafe
 Two Conductors in Parallel (Balanced)
Different Heights Above Ground Plane

For D << d, h1, h2

Z0= (276/ε½)log10{(2D/d)[1+(D2/4h1h2)]-½}

Parallel conductors different heights above a ground plane - RF Cafe
Single Conductor Between
Parallel Ground Planes

For d/h << 0.75

Z0= (138/ε½) log10(4h/πd)

Single conductor between parallel ground planes - RF Cafe

 Two Conductors in Parallel (Balanced)
Between Parallel Ground Planes

For d << D, h

Z0= (276/ε½) log10{[4h tanh(πD/2h)]/πd}

Two conductors between parallel ground planes - RF Cafe
Balanced Conductors Between
Parallel Ground Planes

For d << h

Z0= (276/ε½) log10(2h/πd)

Balanced conductors between parallel ground planes - RF Cafe

 Two Conductors in Parallel (Balanced)
of Unequal Diameters




Z0= (60/ε½) cosh-1 (N)

N = ½[(4D2/d1d2) - (d1/d2) - (d2/d1)]

Parallel conductors - unequal diameters - RF Cafe
Balanced 4-Wire Array

For d << D1, D2

Z0= (138/ε½) log10{(2D2/d)[1+(D2/D1)2]-½}

Balanced 4-wire array - RF Cafe
 Two Conductors
in Open Air


Z0= 276 log10(2D/d)

Two conductors in open air - RF Cafe
5-Wire Array

For d << D

Z0= (173/ε½) log10(D/0.933d)

5-wire array - RF Cafe
 Single Conductor in
Square Conductive Enclosure

For d << D

Z0≈ [138 log10(ρ) +6.48-2.34A-0.48B-0.12C]/ε½

A = (1+0.405ρ-4)/(1-0.405ρ-4)

B = (1+0.163ρ-8)/(1-0.163ρ-8)

C = (1+0.067ρ-12)/(1-0.067ρ-12)

ρ= D/d

Single conductor in square conducting enclosure - RF Cafe
Air Coaxial Cable with
Dielectric Supporting Wedge

For d << D

Z0≈ [138 log10(D/d)]/[1+(ε-1)(θ/360)]½)

ε = wedge dielectric constant

θ= wedge angle in degrees

Air coaxial cable with dielectric supporting wedge - RF Cafe
 Two Conductors Inside Shield
(sheath return)

For d << D, h

Z0= (69/ε½) log10[(ν/2σ2)(1-σ4)]

ν = h/d       σ = h/D

Twin conductors inside shield - RF Cafe
Balanced Shielded Line

For D>>d, h>>d

Z0= (276/ε½) log10{2ν[(1-σ2)/(1+σ2)]}

ν = h/d       σ = h/D

Balanced shielded line equation - RF Cafe
  
Two Conductors in Parallel (Unbalanced)
Inside Rectangular Enclosure

For d << D, h, w

                          ∞
Z0= (276/ε½) {log10[(4h tanh(πD/2h)/πd)- ∑ log10[(1+μm2)/(1-νm2)]}
                            m=1

μm=sinh(πD/2h)/cosh(mπw/2h)

νm=sinh(πD/2h)/sinh(mπw/2h)

Balanced 2-conductor line inside rectangular enclosure - RF Cafe

 

 

Equations appear in "Reference Data for Engineers," Sams Publishing 1993
Windfreak Technologies withwave microwave devices - RF Cafe
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